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જો $0 \leq \theta \leq 2 \pi$ માટે $\mathrm{A}=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$ હોય, તો
$\operatorname{Det}(\mathrm{A})=0$
$\operatorname{Det}(\mathrm{A}) \in[2,4]$
$Det$ $(\mathrm{A}) \in(2, \infty)$
$\operatorname{Det}(\mathrm{A}) \in(2,4)$
Solution
$A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$
$\therefore|A|=1\left(1+\sin ^{2} \theta\right)-\sin \theta(-\sin \theta+\sin \theta)+1\left(\sin ^{2} \theta+1\right)$
$=1+\sin ^{2} \theta+\sin ^{2} \theta+1$
$=2+2 \sin ^{2} \theta$
$=2\left(1+\sin ^{2} \theta\right)$
Now, $0 \leq \theta \leq 2 \pi$
$\Rightarrow 0 \leq \sin \theta \leq 1$
$\Rightarrow 0 \leq \sin ^{2} \theta \leq 1$
$\Rightarrow 1 \leq 1+\sin ^{2} \theta \leq 2$
$\Rightarrow 2 \leq 2\left(1+\sin ^{2} \theta\right) \leq 4$
$\therefore \operatorname{Det}(A) \in[2,4]$
Hence, the correct answer is $B$.
